Question Number 207717 by hardmath last updated on 24/May/24
$$\mathrm{lg}^{\mathrm{2}} \:\left(\mathrm{10x}\right)\:−\:\mathrm{lg}\:\mathrm{10x}\:+\:\mathrm{1}\:=\:\mathrm{6}\:−\:\mathrm{lg}\:\mathrm{x} \\ $$$$\mathrm{find}:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$
Answered by TonyCWX08 last updated on 24/May/24
$$\left(\mathrm{1}+{lg}\left({x}\right)\right)^{\mathrm{2}} −\left(\mathrm{1}+{lg}\left({x}\right)\right)+{lg}\left({x}\right)+\mathrm{1}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{2}{lg}\left({x}\right)+{lg}^{\mathrm{2}} \left({x}\right)−\mathrm{1}−{lg}\left({x}\right)+{lg}\left({x}\right)−\mathrm{5}=\mathrm{0} \\ $$$${lg}^{\mathrm{2}} \left({x}\right)+\mathrm{2}{lg}\left({x}\right)−\mathrm{5}=\mathrm{0} \\ $$$${let}\:{y}\:=\:{lg}\left({x}\right) \\ $$$${y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{5}=\mathrm{0} \\ $$$${y}=−\mathrm{1}\pm\sqrt{\mathrm{6}} \\ $$$${lg}\left({x}\right)=−\mathrm{1}\pm\sqrt{\mathrm{6}} \\ $$$${x}=\mathrm{10}^{−\mathrm{1}\pm\sqrt{\mathrm{6}}} \\ $$$$ \\ $$$${x}_{\mathrm{1}} =\mathrm{10}^{−\mathrm{1}+\sqrt{\mathrm{6}}} \\ $$$${x}_{\mathrm{2}} =\mathrm{10}^{−\mathrm{1}−\sqrt{\mathrm{6}}} \\ $$
Commented by hardmath last updated on 24/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$